probability problem

Suppose that there are $\displaystyle 2 $ teams: $\displaystyle A $ and $\displaystyle B $. Team $\displaystyle A $'s pitcher throws a strike $\displaystyle 50 \% $ of the time and a ball $\displaystyle 50 \% $ of the time (successive pitches are independent from one another), and the pitcher never hits the batter. Knowing this, Team $\displaystyle B $'s manager instructs the first batter not to swing at anything. Calculate the following probabilities:

In order for the batter to be walked on the sixth pitch it must be true that among the first five pitches there must be two strikes and three balls while the sixth pitch must be a ball. Here is an example $\displaystyle \underbrace {BSBSB}_5\boxed{B}$.
But the first five places can be rearranged in $\displaystyle {5 \choose 3} $ ways.
Thus the probability is $\displaystyle {5 \choose 3} \left( {\frac{1}{2}} \right)^6 $.